# Irreducible polynomials in Z[X]

could you please confirm my method of solution for the following question? Thanks in advance!! :)

Consider the polynomial $f(x) = x^3 + x^2 - 2x + 1 \in \mathbb Z[X]$. Prove that $f$ is irreducible in $\mathbb Q[X]$.

Solution: I tried to use Einstein's Criterion but it clearly does not work here. So I tried to solve for $x$ by using the Tschirnaus Transformation and turn it into a compress cubic and solve for $x$. I got $x_1, x_2, x_3 =$ some number that consists of nested square roots and the complex number i.

In this case, isn't it sufficient to show that $f$ is irreducible since the solution $x_1, x_2, x_3$ are clearly not in the set of Rationals?

Thanks! :)

• Excluding integer / rational roots is a bit easier than finding all the roots. Have you heard about the rational root theorem? Commented May 14, 2017 at 21:18
• A degree $3$ polynomial over $\mathbb{Q}$ is reducible if and only if it has a root in $\mathbb{Q}$. Commented May 14, 2017 at 21:20
• @Arthur Not really. I am looking it up right now. :( Commented May 14, 2017 at 21:24
• @Arthur Do you think my way is still correct? Because this is from the final exam, and the prof marked it as incorrect and awarded me a 0 out of 5. I am planning to petition it. Commented May 14, 2017 at 21:25
• @Wilson the solution x_1, x_2, x_3 are clearly not in the set of Rationals It's not enough to say clearly, you need to prove that. Just because an expression contains radicals, it doesn't mean it can't be rational, and just because it contains $i$ it doesn't mean it can't be real.
– dxiv
Commented May 14, 2017 at 23:04

If $\frac{p}{q}$ is a rational solution of $x^3+x^2-2x+1$ with $p,q\in \mathbb{Z}$ and $p,q$ coprime then
$$\frac{p^3}{q^3}+\frac{p^2}{q^2}-2\frac{p}{q}+1=0 \Leftrightarrow p^3+p^2q-2pq^2+q^3 = 0\Leftrightarrow p(p^2+pq-2q^2) = -q^3.$$ This implies that $p|q$. But since $p$ and $q$ are coprime it follows $p=\pm 1$. In the same way
$$\frac{p^3}{q^3}+\frac{p^2}{q^2}-2\frac{p}{q}+1=0 \Leftrightarrow p^3+p^2q-2pq^2+q^3 = 0\Leftrightarrow -p^3 = q(p^2-2pq+q^2).$$ This implies $q|p$. But since since $p$ and $q$ are coprime it follows $q=\pm 1$.
Hence, if there is a rational root of $x^3+x^2-2x+1$ it must be $\pm 1$ which is obviously not a solution. Consequently there is no rational number $\frac{p}{q}$ such that $(x-\frac{p}{q})\mid ( x^3+x^2-2x+1)$ which proves that $x^3+x^2-2x+1$ is irreducible.